Упр.1036 ГДЗ Алимов 10-11 класс (Алгебра)

1036 1) интеграл (0;1) (5×4-8×3)dx;
2) интеграл (-1;1=2) (6×3-5x)dx;
3) интеграл (1;4) корень x(3-7/x)dx;
4) интеграл (1;8) 4 корень 3 степени x(1-4/x)dx;
5) интеграл (0;3) корень (x+1)dx;
6) интеграл (2;6) корень(2x-3)dx.

1. $$\int_0^1 (5x^4-8x^3)\,dx=\left(x^5-2x^4\right)\Big|_0^1=1-2=-1.$$

2. $$\int_{-1}^2 (6x^3-5x)\,dx=\left(\frac{3x^4}{2}-\frac{5x^2}{2}\right)\Big|_{-1}^2$$
   $$=\left(24-10\right)-\left(\frac{3}{2}-\frac{5}{2}\right)=14-(-1)=15.$$

3. $$\int_1^4 \sqrt{x}\left(3-\frac{7}{x}\right)\,dx=\int_1^4 \left(3x^{1/2}-7x^{-1/2}\right)\,dx$$
   $$=\left(2x\sqrt{x}-14\sqrt{x}\right)\Big|_1^4$$
   $$=(16-28)-(2-14)=0.$$

4. $$\int_1^8 4\sqrt[3]{x}\left(1-\frac{4}{x}\right)\,dx=\int_1^8 \left(4x^{1/3}-16x^{-2/3}\right)\,dx$$
   $$=\left(3x\sqrt[3]{x}-48\sqrt[3]{x}\right)\Big|_1^8$$
   $$=(24\cdot 2-48\cdot 2)-(3-48)=-3.$$

5. $$\int_0^3 \sqrt{x+1}\,dx=\int_0^3 (x+1)^{1/2}\,dx$$
   $$=\frac{2}{3}(x+1)^{3/2}\Big|_0^3=\frac{2}{3}\left(4^{3/2}-1^{3/2}\right)$$
   $$=\frac{2}{3}(8-1)=\frac{14}{3}=4\frac{2}{3}.$$

6. $$\int_2^6 \sqrt{2x-3}\,dx=\int_2^6 (2x-3)^{1/2}\,dx$$
   $$=\frac{1}{3}(2x-3)^{3/2}\Big|_2^6=\frac{1}{3}\left(9^{3/2}-1^{3/2}\right)$$
   $$=\frac{1}{3}(27-1)=\frac{26}{3}=8\frac{2}{3}.$$

Ответ

1) $$-1$$; 2) $$15$$; 3) $$0$$; 4) $$-3$$; 5) $$\frac{14}{3}$$; 6) $$\frac{26}{3}$$.